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Let $f:V_1 \to V_2$ and $g: V_2 \to V_3$ be two homomorphisms, $f$ is injective and $g$ is surjective, $gf =0$ [I think it is $g(f)$]. Proof that $dimV_2 \ge dimV_1 + dimV_3$.

$ V_1,V_2,V_3$ are the vector spaces.

My attempt:

Because $g$ is an epimorphism (surjective), $V_3$ must be a zero space if $gf$ = 0, right?

Then we have just $dimV_2 \ge dimV_1$ . Because $f$ is injective, $V_2$ must contain whole $V_1$ so $V_2$ must be at least as big as $V_1$.

I think I am wrong, especially with $dimV_3 = 0$. Thanks for the help.

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    It's not $V_3$=0, but $\text{im}(g\circ f)=0$. – 2017-01-31
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    Simple counterexample(of $V_3=0$): $V_1=V_3=\Re$, $V_2=\Re^2$, $f: x \mapsto (x,0)$, $g: (x,y) \mapsto y$. – 2017-01-31
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    Ok thank you, I see it right now. – 2017-01-31

1 Answers 1

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I hope all vector spaces are finite-dimensional. We use the dimension theorem for linear maps. Since $f$ is injective, $\dim(V_1)=\dim(\text{im}(f))$. Since $g$ is surjective, $\dim(V_3)=\dim(V_2)-\dim(\ker(g))$. Notice that $g\circ f:V_1\rightarrow V_3$ and since $g\circ f=0$, we have that $\dim(V_1)=\dim(\ker(g\circ f))$. Notice that $\dim(\ker(g\circ f))\leq \dim(\ker(g))$ (Why?). Putting these results together, we have

$$\dim(V_1)\leq \dim(\ker(g))=\dim(V_2)-\dim(V_3),$$ which we needed to show.

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    because $ ker(f) = 0$ and $g$ is an epimorphism? – 2017-01-31
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    The kernel of $g(f)$ is in $V_1$, right? – 2017-01-31
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    I draw some images but I can't imagine it properly. – 2017-01-31
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    My last comment was wrong. But think about this: We know that $g\circ f=0$. Hence $\text{im}(f)\subset \ker(g)$, can you take it from there? – 2017-01-31
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    So I can separetaly solve $f$ and $g$ as monomorphism and epimorphism and then work with that linear mapping from $V_1$ to $V_3$ where $V_1$ is the place where $Ker(g(f))$ lies, because $Ker(g(f)) \to Im(g(f)) = 0$. I still don't understand why $Ker(g(f))$ is smaller than $Ker(g)$, but thank you very much for your help. :) – 2017-01-31
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    Just think about it. If you look at $\ker(g\circ f)$ then you only look at those things that get mapped to zero by $g$ that lie in the image of $f$. Whereas $\ker(g)$ looks at everything mapped to zero by $g$. – 2017-01-31
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    Yeah, that sounds easy, but how could I prove it? I am sorry I am bothering for so long time, but I am having exam tomorrow. – 2017-01-31
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    Consider the map $\phi:\ker(g\circ f)\rightarrow \ker(g):x\mapsto f(x)$, clearly this map is well-defined and injective, the result follows. – 2017-01-31
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    So, we can also say that $Ker(g(f)) \ge Ker(f)$ because if $dimKer(f) = dimV_1$ then $dimIm(f)=0$ and $dimKer(f)=dimKer(g(f))$, and if $Ker(f)$ would be very small, then $Im(f)$ could be big and $Ker(g)$ is in it, so $Ker(g(f))$ would be larger than $Ker(f)$. Horribly written but I think I understand it now. – 2017-01-31